JUDKINS'S CATTLE.
(
Money Puzzles)
Hiram B. Judkins, a cattle-dealer of Texas, had five droves of animals,
consisting of oxen, pigs, and sheep, with the same number of animals in
each drove. One morning he sold all that he had to eight dealers. Each
dealer bought the same number of animals, paying seventeen dollars for
each ox, four dollars for each pig, and two dollars for each sheep; and
Hiram received in all three hundred and one dollars. What is the
greatest number of animals he could have had? And how many would there
be of each kind?
Answer:
As there were five droves with an equal number of animals in each drove,
the number must be divisible by 5; and as every one of the eight dealers
bought the same number of animals, the number must be divisible by 8.
Therefore the number must be a multiple of 40. The highest possible
multiple of 40 that will work will be found to be 120, and this number
could be made up in one of two ways--1 ox, 23 pigs, and 96 sheep, or 3
oxen, 8 pigs, and 109 sheep. But the first is excluded by the statement
that the animals consisted of "oxen, pigs, and sheep," because a single
ox is not oxen. Therefore the second grouping is the correct answer.
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